Question

The attractive electrostatic force between the point charges 6.14×10−6 c and q has a magnitude of 0.845 n when the separation between the charges is 9.26 m . part a find the sign and magnitude of the charge q.

Answer 1

Coulomb's law states that F = k(q1)(q2)/r^2, where k = 9 x 10^9 N-m^2/C^2
Substituting the given values:

0.845 N = (9x10^9 N-m^2/C^2)(6.14x10^-6 C)(q) / (9.26^2)
q = 0.00131 = 1.31 x 10^-3 C

Since the force is attractive, the point charges must be of opposite signs, and q is negatively charged (-1.31 x 10^-3 C).